**1 Answer**

written 3.9 years ago by |

**The connecting rod is $1.2m$ long. When the crank has turned through $125°$ from the TDC, The steam pressure above the piston is $30 KN/m^2$ and below the piston is $1.5 KN/m^2$, Calculate the effective turning moment on the crankshaft.**

$\rightarrow$ **Given:**

$D = 300 mm$

Stroke $= L = 450 mm$

$\therefore$ Radius of crank $= r = \frac{L}{2} = \frac{450}{2} = 225 \ mm$

$M_R = 225 kg$

Piston rod dia. $= d = 50 mm$

$l = 1.2 m$

$\theta = 125°$

$P_1 = 30 \ KN/m^2$

$P_2 = 1.5 \ KN/m^2$

$A_1 \ = \ \frac{\pi}{4} (0.300)^2 \ = \ 0.0706 \ m^2$

$a \ = \ \frac{\pi}{4} (0.050)^2 \ = \ 1.963 \times \ 10^{-3} \ m^2$

$N = 200 rpm$

$\omega = 20.94 rad/s$

$\rightarrow$ $F_L = P_1 A_1 – P_2 [ A_1 – a]$$= 30 \times 0.0706 – 1.5 [0.0706 – 1.963 \times 10^{-3}]$ $F_L = 2.015 \ KN$ $\rightarrow$ $n = \frac{l}{r} = \frac{1.2}{0.225} = 5.33$

$\rightarrow$ $F_I \ = \ m_R. \ w^2.r. \ [cos \ \theta + \frac{cos \ 2 \ \theta}{n}]$

$= (225) . (20.94)^2 . \ (0.225) \ [ cos \ 125 + \frac{cos \ (2 \times 125)}{5.33}]$

$= -14.16 \times 10^3 \ N$

$\rightarrow$ **Net force on piston or piston effort,**

$F_p = F_L – F_I + M_R . g$ $\rightarrow$ As engine is vertical type therefore height of reciprocating is to be considered.

$= 2.015 – (-14.16) + ( \frac{225 \times 981}{1000)})$

$= 18.38 KN$

$\rightarrow$ $sin \ \phi \ = \frac{sin \ \theta}{n} = \frac{sin \ (125)}{5.33} = 0.1536$

$\phi = 8.84°$

$\rightarrow$ **Effective turning moment on crank shaft.**

$T = F_T . r$

$= F_Q.sin \ (\theta \ + \ \phi) \ . r$

$= (\frac{FP}{cos \ \theta}). sin \ (\theta \ + \ \phi). r$

$= (\frac{18.38 \times 10^3}{cos \ 8.84}) . sin \ (125 + 8.84) (0.225)$

$= 3.0186 \times 10^3 \ N-m$